g An electron enters a region of space containing a uniform 1.63 × 10 − 5 T magnetic field. Its speed is 121 m/s and it enters p
2 answers:
<h2>Answer:</h2><h2></h2>
(a) 422.2 x 10⁻⁷m
(b) 4.6 x 10⁵ Hz
<h2>Explanation:</h2>
The magnetic force acting perpendicular to the velocity of the electron will cause a circular motion so that the centripetal force, , of the electron is equal to the Lorentz's force,
. i.e
=
---------------(i)
Where;
=
= qvB
Equation (i) then becomes;
= qvB ------------------(ii)
Where;
m = mass of the electron
v = linear velocity of the electron
r = radius of the electron's path
q = charge of the electron
B = magnetic field.
Make r subject of the formula in equation (ii)
r =
r = -----------(iii)
From the question;
v = 121m/s
B = 1.63 x 10⁻⁵ T
q = 1.6 x 10⁻¹⁹ C (known constant)
m = 9.1 x 10⁻³¹kg
Substitute these values into equation (iii) as follows;
r =
r = 422.2 x 10⁻⁷m
Therefore, the radius of the electron's path is 422.2 x 10⁻⁷m
(ii) The frequency, f, of the motion which is also called the cyclotron frequency - the number of cycles the electron completes around the path every second - is given by;
f =
Substitute the values of v and r into the equation as follows;
f =
Take = 3.142
f =
f = 4.6 x 10⁵ Hz
Therefore, the frequency of the motion is 4.6 x 10⁵ Hz
Answer:
i. The radius 'r' of the electron's path is 4.23 × m.
ii. The frequency 'f' of the motion is 455.44 KHz.
Explanation:
The radius 'r' of the electron's path is called a gyroradius. Gyroradius is the radius of the circular motion of a charged particle in the presence of a uniform magnetic field.
r =
Where: B is the strength magnetic field, q is the charge, v is its velocity and m is the mass of the particle.
From the question, B = 1.63 × T, v = 121 m/s, Θ =
(since it enters perpendicularly to the field), q = e = 1.6 ×
C and m = 9.11 ×
Kg.
Thus,
r = ÷ sinΘ
But, sinΘ = sin = 1.
So that;
r =
= (9.11 × × 121) ÷ (1.6 ×
× 1.63 ×
)
= 1.10231 × ÷ 2.608 ×
= 4.2266 ×
= 4.23 × m
The radius 'r' of the electron's path is 4.23 × m.
B. The frequency 'f' of the motion is called cyclotron frequency;
f =
= (1.6 × × 1.63 ×
) ÷ (2 ×
× 9.11 ×
)
= 2.608 × ÷ 5.7263 ×
= 455442.4323
f = 455.44 KHz
The frequency 'f' of the motion is 455.44 KHz.
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Answer:
<em>a) 6738.27 J</em>
<em>b) 61.908 J</em>
<em>c) </em>
<em></em>
Explanation:
The complete question is
A flywheel is a mechanical device used to store rotational kinetic energy for later use. Consider a flywheel in the form of a uniform solid cylinder rotating around its axis, with moment of inertia I = 1/2 mr2.
Part (a) If such a flywheel of radius r1 = 1.1 m and mass m1 = 11 kg can spin at a maximum speed of v = 35 m/s at its rim, calculate the maximum amount of energy, in joules, that this flywheel can store?
Part (b) Consider a scenario in which the flywheel described in part (a) (r1 = 1.1 m, mass m1 = 11 kg, v = 35 m/s at the rim) is spinning freely at its maximum speed, when a second flywheel of radius r2 = 2.8 m and mass m2 = 16 kg is coaxially dropped from rest onto it and sticks to it, so that they then rotate together as a single body. Calculate the energy, in joules, that is now stored in the wheel?
Part (c) Return now to the flywheel of part (a), with mass m1, radius r1, and speed v at its rim. Imagine the flywheel delivers one third of its stored kinetic energy to car, initially at rest, leaving it with a speed vcar. Enter an expression for the mass of the car, in terms of the quantities defined here.
moment of inertia is given as
=
where m is the mass of the flywheel,
and r is the radius of the flywheel
for the flywheel with radius 1.1 m
and mass 11 kg
moment of inertia will be
=
= 6.655 kg-m^2
The maximum speed of the flywheel = 35 m/s
we know that v = ωr
where v is the linear speed = 35 m/s
ω = angular speed
r = radius
therefore,
ω = v/r = 35/1.1 = 31.82 rad/s
maximum rotational energy of the flywheel will be
E = = 6.655 x
= <em>6738.27 J</em>
<em></em>
b) second flywheel has
radius = 2.8 m
mass = 16 kg
moment of inertia is
=
=
= 62.72 kg-m^2
According to conservation of angular momentum, the total initial angular momentum of the first flywheel, must be equal to the total final angular momentum of the combination two flywheels
for the first flywheel, rotational momentum = = 6.655 x 31.82 = 211.76 kg-m^2-rad/s
for their combination, the rotational momentum is
where the subscripts 1 and 2 indicates the values first and second flywheels
= (6.655 + 62.72)ω
where ω here is their final angular momentum together
==> 69.375ω
Equating the two rotational momenta, we have
211.76 = 69.375ω
ω = 211.76/69.375 = 3.05 rad/s
Therefore, the energy stored in the first flywheel in this situation is
E = = 6.655 x
= <em>61.908 J</em>
<em></em>
<em></em>
c) one third of the initial energy of the flywheel is
6738.27/3 = 2246.09 J
For the car, the kinetic energy =
where m is the mass of the car
is the velocity of the car
Equating the energy
2246.09 =
making m the subject of the formula
mass of the car m =
Answer:
Total distance = 700 m
Displacement = 500 m
Explanation:
Notice that Jed travelled a total of 3 x 100 m = 300 m in the North direction, and 300 m + 100 m = 400 m in the East direction. Therefore the total distance he travelled is: 300 + 400 = 700 m.
But the actual displacement is given by the Pythagorean theorem as the hypotenuse of a right angle triangle of legs 300 m and 400 m:
displacement =